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8/14/2019 CHAP_3_SEC1.PDF
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57:020 Mechanics of Fluids and Transport Processes Chapter 3Professor Fred Stern Typed by Stephanie Schrader Fall 1999 1
Chapter 3: Fluid Statics
3.1 Pressure
For a static fluid, the only stress is the normal stress since by definition a fluid subjected to a shear stress must deformand undergo motion. Normal stresses are referred to as
pressure p.
For the general case, the stress on a fluid element or at a point is a tensor
For a static fluid,ij = 0 i j i.e., shear stresses = 0
ii = p = xx = yy = zz i = j i.e., normal stresses= p
Also shows that p is isotropic i.e., one value at a pointwhich is independent of direction, a scalar.
ij = stress tensor
= xx xy xzyx yy yz
zx zy zz
i = force= direction
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57:020 Mechanics of Fluids and Transport Processes Chapter 3Professor Fred Stern Typed by Stephanie Schrader Fall 1999 2
x z
Definition of Pressure:
dA
dF
A
F
0A
lim p =
=
F = normal force acting over A N/m 2 = Pa (Pascal)
As already noted, p is a scalar, which can be easilydemonstrated by considering the equilibrium of forces on awedge-shaped fluid element
GeometryA = lyx = lcos z = lsin
Fx = 0 pnA sin - p xA sin = 0 pn = p x
Fz = 0-pnA cos + p zA cos - W = 0
y)sin)(cos(2
W = ll
pnA
pxAsin
W=wei ht pzAcos
x
l
z
W = mg= Vg= V
V = xzy
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57:020 Mechanics of Fluids and Transport Processes Chapter 3Professor Fred Stern Typed by Stephanie Schrader Fall 1999 3
+ = p pn z 2 0l sin
p p for n z= l 0 i.e., p n = p x = p y = p z
p is single valued at a point and independent of direction
A body/surface in contact with a static fluid experiences aforce due to p
=BS
p dAn pF
Note: if p = constant, F p = 0 for a closed body
Scalar form of Greene's Theorem:f nds fdV
s V =
f = constant f = 0
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57:020 Mechanics of Fluids and Transport Processes Chapter 3Professor Fred Stern Typed by Stephanie Schrader Fall 1999 4
Pressure Transmission
Pascal's law: in a closed system, a pressure change
produced at one point in the system is transmittedthroughout the entire system.
Absolute Pressure, Gage Pressure, and Vacuum
For p A>p a, pg = p A p a = gage pressure
For p A
0
pA > p a pa = atmospheric
pressure =101.325 kPa
pA = 0 = absolutezero
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57:020 Mechanics of Fluids and Transport Processes Chapter 3Professor Fred Stern Typed by Stephanie Schrader Fall 1999 5
3.2 Pressure Variation with Elevation
Basic Differential Equation
For a static fluid, pressure varies only with elevation withinthe fluid. This can be shown by consideration ofequilibrium of forces on a fluid element
Newton's law (momentum principle) applied to a staticfluid
F = ma = 0 for a static fluidi.e., Fx = Fy = Fz = 0
Fz = 0
pdxdy p pz dz dxdy gdxdydz + =( )
0
pz
g= =
Basic equation for pressure variation with elevation
1st order Taylor series
estimate for pressurevariation over dz
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57:020 Mechanics of Fluids and Transport Processes Chapter 3Professor Fred Stern Typed by Stephanie Schrader Fall 1999 6
0y
p
0dxdz)dyy
p p( pdxdz
0Fy
=
=+
=
0x
p
0dydz)dxx p
p( pdydz
0Fx
=
=+
=
For a static fluid, the pressure only varies with elevation zand is constant in horizontal xy planes.
The basic equation for pressure variation with elevation can be integrated depending on whether = constant or = (z), i.e., whether the fluid is incompressible (liquid orlow-speed gas) or compressible (high-speed gas) sinceg constant
Pressure Variation for a Uniform-Density Fluid= g = constant
dpdz
=
dp = dz p = z + constant p + z = constant
piezometric pressure
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57:020 Mechanics of Fluids and Transport Processes Chapter 3Professor Fred Stern Typed by Stephanie Schrader Fall 1999 7
Alternate forms:
p1 + z1 = p 2 + z2
p = - z
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57:020 Mechanics of Fluids and Transport Processes Chapter 3Professor Fred Stern Typed by Stephanie Schrader Fall 1999 8
Pressure Variation for Compressible Fluids:
Basic equation for pressure variation with elevation
g)z, p(dzdp = = =
Pressure variation equation can be integrated for (p,z)known. For example, here we solve for the pressure in theatmosphere assuming (p,T) given from ideal gas law, T(z)known, and g g(z).
p = RT R = gas constant = 287 J/kg K p,T in absolute scale
RT pg
dzdp =
)z(Tdz
R g
pdp
= which can be integrated for T(z) known
dry air
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57:020 Mechanics of Fluids and Transport Processes Chapter 3Professor Fred Stern Typed by Stephanie Schrader Fall 1999 9
zo = earth surface= 0
po = 101.3 kPa
T = 15C
= 6.5 K/km
Pressure Variation in the Troposphere
T = T o (z z o) linear decrease
To = T(z o) where p = p o(zo) known = lapse rate = 6.5 K/km
)]zz(T[dz
R g
pdp
oo =
dz'dz
)zz(T'z oo=
=
constant)]zz(Tln[R g pln oo +=
use reference condition
constantTlnR g
pln oo +=
solve for constant
R g
o
oo
o
o
oo
o
T)zz(T
p p
T)zz(T
lnR
g p p
ln
=
=
i.e., p decreases for increasing z
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57:020 Mechanics of Fluids and Transport Processes Chapter 3Professor Fred Stern Typed by Stephanie Schrader Fall 1999 10
Pressure Variation in the Stratosphere
T = T s = 55C
dp p
gR
dzTs
=
constantzRT
g pln
s+=
use reference condition to find constant
]RT/g)zz(exp[ p p
e p p
soo
RT/g)zz(
o
s0
=
=
i.e., p decreases exponentially for increasing z.
See examples 3.5 and 3.6 in text for p analysis inTroposphere and Stratosphere, respectively.
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57:020 Mechanics of Fluids and Transport Processes Chapter 3Professor Fred Stern Typed by Stephanie Schrader Fall 1999 11
3.3 Pressure Measurements
Pressure is an important variable in fluid mechanics and
many instruments have been devised for its measurement.Many devices are based on hydrostatics such as barometersand manometers, i.e., determine pressure throughmeasurement of a column (or columns) of a liquid usingthe pressure variation with elevation equation for anincompressible fluid.
More modern devices include Bourdon -Tube Gage(mechanical device based on deflection of a spring) and
pressure transducers (based on deflection of a flexiblediaphragm/membrane). The deflection can be monitored
by a strain gage such that voltage output is p acrossdiaphragm, which enables electronic data acquisition withcomputers.
In this course we will use both manometers and pressuretransducers in EFD labs 2 and 3.
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57:020 Mechanics of Fluids and Transport Processes Chapter 3Professor Fred Stern Typed by Stephanie Schrader Fall 1999 12
Manometry
1. Barometer
pv + Hgh = p atm
i.e., p atm = Hgh p v 0 i.e., vapor pressure Hgnearly zero at normal T
h 76 cm patm 101 kPa (or 14.6 psia)
Note: p atm is relative to absolute zero, i.e., absolute pressure. p atm = p atm(location, weather)
Consider why water barometer is impracticalOHOHHgHg 22 hh =
.ft34cm6.1033766.13hShh HgHgHgOH
HgOH
2
2==== =
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57:020 Mechanics of Fluids and Transport Processes Chapter 3Professor Fred Stern Typed by Stephanie Schrader Fall 1999 13
patm
2. Piezometer
patm + h = p pipe = p absolute
p = h gage
Simple but impractical for large p and vacuum pressures(i.e., p abs < p atm)
3. U-tube or differential manometer
p1 + mh l = p 4 p1 = p atm p4 = mh l gage
= w[Smh S l ]
for gases S
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57:020 Mechanics of Fluids and Transport Processes Chapter 3Professor Fred Stern Typed by Stephanie Schrader Fall 1999 14
Air at 20 C is in pipe with a water manometer. For givenconditions compute gage pressure in pipe.
l = 140 cmh = 70 cm
p4 = ? gage (i.e., p 1 = 0)
p1 + h = p 3 step-by-step method p3 - air l = p 4 p1 + h - air l = p 4 complete circuit method
h - air l = p 4 gage
water (20C) = 9790 N/m 3 p3 = h = 6853 Pa [N/m 2] air = g
pabs( )( )
3atm3 m/kg286.1)27320(287
1013006853273CR
p pRT p =+
+=++==
K air = 1.286 9.81m/s 2 = 12.62 N/m 3
note air
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57:020 Mechanics of Fluids and Transport Processes Chapter 3Professor Fred Stern Typed by Stephanie Schrader Fall 1999 15
A differential manometer determines the difference in pressures at two points and when the actual pressureat any point in the system cannot be determined.
h)()( p p
p)h(h p
f m12f 21
22f m1f 1
+ =
=+
ll
ll
h1 p pf
m2
f
21
f
1
=
+
+
ll
difference in piezometric head
if fluid is a gas f